Question: Evaluate the improper integral if it exists. $\int^{0}_{-\infty}e^{x}\,dx $ Choose 1 answer: Choose 1 answer: (Choice A) A $-1$ (Choice B) B $0$ (Choice C) C $1$ (Choice D) D The improper integral diverges.
Answer: First, let's rewrite the improper integral: $\int^{0}_{-\infty}e^{x}\,dx =\lim_{a\to-\infty}\int_a^0 e^{x}\,dx$ We can now evaluate the integral: $\begin{aligned} \phantom{\int_{-\infty}^{0}e^x\,dx}&=\lim_{a\to -\infty}\int_a^0 e^x\,dx\\ \\ \\ &=\lim_{a\to-\infty}\Big[e^x\Big]_a^0\\ \\ \\ &=\lim_{a\to-\infty}(e^0-e^{a})\\ \\ &=\lim_{a\to-\infty}1-\lim_{a\to-\infty}e^a\\ \\ &=1-0\\ \\ &=1 \end{aligned}$ The answer: $\int^{0}_{-\infty}e^{x}\,dx =1$